Let $f'(x) > 0$ and $g'(x) < 0$ for all $x \in \mathbb{R}$. Then which of the following is true?

The composite mapping $fog$ of the maps $f:R \to R$,$f(x) = \sin x$,and $g:R \to R$,$g(x) = x^2$ is:

Let $f(x) = \sin \left(\frac{\pi}{6} \sin \left(\frac{\pi}{2} \sin x\right)\right)$ for all $x \in R$ and $g(x) = \frac{\pi}{2} \sin x$ for all $x \in R$. Let $(f \circ g)(x)$ denote $f(g(x))$ and $(g \circ f)(x)$ denote $g(f(x))$. Then which of the following is (are) true?
$(A)$ Range of $f$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$(B)$ Range of $f \circ g$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$(C)$ $\lim _{x \rightarrow 0} \frac{f(x)}{g(x)} = \frac{\pi}{6}$
$(D)$ There is an $x \in R$ such that $(g \circ f)(x) = 1$

Let $f: \{1,3,4\} \rightarrow \{1,2,5\}$ and $g: \{1,2,5\} \rightarrow \{1,3\}$ be given by $f = \{(1,2), (3,5), (4,1)\}$ and $g = \{(1,3), (2,3), (5,1)\}$. Write down $g \circ f$.

If $f(x) = \frac{1 - x}{1 + x}$, then $f(f(\cos x)) = $

For $x \in \left( 0, \frac{3}{2} \right)$,let $f(x) = \sqrt{x}$,$g(x) = \tan x$,and $h(x) = \frac{1 - x^2}{1 + x^2}$. If $\phi(x) = ((h \circ f) \circ g)(x)$,then $\phi\left( \frac{\pi}{3} \right)$ is equal to